JEE Chemistry Hydrocarbons MCQs Set 11

Answer: (b) 2-Chloro-2-methylpropane and 1,1-dichloroethane

Explanation:
1. Addition to the alkene follows Markovnikov's rule, so Cl goes to the more substituted carbon giving \( \ce{(CH3)3C-Cl} \)
2. Ethyne adds one HCl to give \( \ce{CH2=CHCl} \), and the second HCl again follows Markovnikov's rule to give \( \ce{CH3-CHCl2} \)

Teacher's Note:
1. Electrophilic addition goes through the more stable carbocation, so H adds to the carbon with more hydrogens.
2. With HBr and a peroxide the radical path reverses the orientation.

Answer: (d) Increases

Explanation:
1. A longer chain has a larger surface area
2. This strengthens van der Waals forces, so the boiling point rises

Teacher's Note:
1. Recall the key fact or reaction first, then eliminate options that contradict it.
2. Link each answer to the underlying bonding or mechanism rather than memorising it.

Answer: (d) High boiling constituents condense at the bottom of the column

Explanation:
1. The column is hottest at the bottom and cooler towards the top
2. Heavy, high boiling fractions condense low down, while light fractions rise and condense near the top

Teacher's Note:
1. Straight chain alkanes knock easily, while branched alkanes and aromatics resist knocking.
2. Octane number is the percentage of iso-octane in a mixture with n-heptane.

Answer: (d) The fraction of crude oil that distils between about 40 and 180 degrees C

Explanation:
1. Crude oil is separated by fractional distillation
2. The light liquid fraction boiling roughly between 40 and 180 degrees C is gasoline (petrol)

Teacher's Note:
1. Fractions of crude oil come off in order of boiling point: gases, petroleum ether, gasoline, kerosene, diesel, lubricating oil.
2. Cracking breaks large molecules into smaller, more useful ones.

Answer: (d) High molecular weight organic compounds decompose to give low molecular weight organic compounds

Explanation:
1. Cracking is the thermal or catalytic breaking of large molecules into smaller ones
2. Heavy hydrocarbons give lighter alkanes and alkenes useful as fuels

Teacher's Note:
1. Remember the composition of common fuels: LPG is butane and propane, natural gas is mostly methane, producer gas is CO with N2.
2. Match each fuel to its main constituent.

Answer: (d) n-Heptane

Explanation:
1. The octane scale is defined with n-heptane as 0
2. iso-Octane (2,2,4-trimethylpentane) is defined as 100

Teacher's Note:
1. Straight chain alkanes knock easily, while branched alkanes and aromatics resist knocking.
2. Octane number is the percentage of iso-octane in a mixture with n-heptane.

Answer: (d) Ethane

Explanation:
1. Decarboxylation removes one carbon as carbonate: \( \ce{CH3CH2COONa + NaOH -> CH3CH3 + Na2CO3} \)
2. The alkane has one carbon less than the acid salt

Teacher's Note:
1. Identify the functional group that is removed or formed and count the carbon atoms of the product.
2. Wurtz and Kolbe couple two alkyl groups, soda lime removes one carbon, and Grignard reagents pick up an acidic proton.

Answer: (a)

Explanation:
1. Bromination is highly selective and favours the most stable radical
2. Abstraction of a secondary hydrogen from n-butane gives the secondary radical and hence 2-bromobutane

Teacher's Note:
1. Draw the possible intermediates and pick the most stable one (tertiary, then secondary, then primary).
2. Only HBr shows the peroxide effect, not HCl or HI.

Answer: (c) 200 - 300

Explanation:
1. Fractions by boiling range: petroleum ether and gasoline up to about 200, kerosene about 200 to 300
2. Above 300 come diesel, lubricating oil and residue

Teacher's Note:
1. Remember the composition of common fuels: LPG is butane and propane, natural gas is mostly methane, producer gas is CO with N2.
2. Match each fuel to its main constituent.

Answer: (d) Are all straight chain compounds

Explanation:
1. Alkanes include branched members such as isobutane and neopentane
2. So the statement that all are straight chain is wrong

Teacher's Note:
1. Check the type of reaction involved and what it does to the carbon skeleton.
2. Compare the options one by one against that.

Answer: (d) Antiknock property of the fuel increases

Explanation:
1. Tetraethyl lead slows the chain reactions that cause pre-ignition
2. It therefore raises the octane number and reduces knocking

Teacher's Note:
1. Write the balanced equation first and track every carbon.
2. Check whether the reagent gives elimination (alcoholic KOH) or substitution (aqueous KOH).

Answer: (d) Cracking

Explanation:
1. Cracking breaks large molecules into small ones
2. Heavy liquid hydrocarbons give gaseous alkanes and alkenes

Teacher's Note:
1. Straight chain alkanes knock easily, while branched alkanes and aromatics resist knocking.
2. Octane number is the percentage of iso-octane in a mixture with n-heptane.

Answer: (d) Le Bell and van't Hoff

Explanation:
1. Le Bell and van't Hoff independently proposed the tetrahedral carbon atom in 1874
2. Kekule gave the structure of benzene and Pauling hybridisation

Teacher's Note:
1. Aromatic rings keep their stability by electrophilic substitution, not addition.
2. Activating groups direct ortho and para, while strongly withdrawing groups direct meta.

Answer: (d) \( \ce{CH3CH2OH} \)

Explanation:
1. A Grignard reagent \( \ce{RMgX} \) takes a proton from an acidic O-H and gives \( \ce{R-H} \), an alkane
2. Aldehydes, ketones and esters instead give alcohols by addition

Teacher's Note:
1. Identify the functional group that is removed or formed and count the carbon atoms of the product.
2. Wurtz and Kolbe couple two alkyl groups, soda lime removes one carbon, and Grignard reagents pick up an acidic proton.

Answer: (d) n-Pentane

Explanation:
1. The first four alkanes are gases
2. n-Pentane (boiling point about 36 degrees C) is a liquid

Teacher's Note:
1. Trends such as boiling point and stability depend on molecular size, branching and substituent effects.
2. Apply the trend directly to rank the options.

Answer: (a) A large number of alkanes are soluble in water

Explanation:
1. Alkanes are non-polar and are not soluble in water
2. So the statement about solubility is false

Teacher's Note:
1. Recall the key fact or reaction first, then eliminate options that contradict it.
2. Link each answer to the underlying bonding or mechanism rather than memorising it.

Answer: (d) Synthetic petrol

Explanation:
1. Water gas (\( \ce{CO + H2} \)) over a catalyst gives a mixture of hydrocarbons
2. This mixture is synthetic petrol

Teacher's Note:
1. Fractions of crude oil come off in order of boiling point: gases, petroleum ether, gasoline, kerosene, diesel, lubricating oil.
2. Cracking breaks large molecules into smaller, more useful ones.

Answer: (d) \( \ce{CH4} \)

Explanation:
1. Wurtz coupling joins two alkyl groups, so the product has at least two carbons
2. Methane has only one carbon and cannot be formed

Teacher's Note:
1. Write the balanced equation first and track every carbon.
2. Check whether the reagent gives elimination (alcoholic KOH) or substitution (aqueous KOH).

Answer: (d) Dehydrohalogenation

Explanation:
1. Sodium ethoxide is a strong base used in ethanol to remove HX from alkyl halides
2. The product is an alkene

Teacher's Note:
1. Cold dilute alkaline KMnO4 gives a diol from an alkene, while hot acidic KMnO4 cleaves the double bond.
2. Terminal alkynes are separated from other hydrocarbons by their acidic hydrogen.

Answer: (d) \( \ce{CH4} \)

Explanation:
1. Percentage of H: methane 4/16 = 25%, ethene 4/28 = 14.3%, benzene 6/78 = 7.7%, ethyne 2/26 = 7.7%
2. Methane is highest

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (d) \( \ce{C3H8} \)

Explanation:
1. Moles = 5.6/22.4 = 0.25
2. Molar mass = 11/0.25 = 44 g/mol
3. \( \ce{CnH2n+2} \) with 14n + 2 = 44 gives n = 3, so \( \ce{C3H8} \)

Teacher's Note:
1. Check the type of reaction involved and what it does to the carbon skeleton.
2. Compare the options one by one against that.

Answer: (b)

Explanation:
1. iso-Octane is 2,2,4-trimethylpentane
2. It has a quaternary carbon at C-2 and a methyl on C-4

Teacher's Note:
1. Straight chain alkanes knock easily, while branched alkanes and aromatics resist knocking.
2. Octane number is the percentage of iso-octane in a mixture with n-heptane.

Answer: (a)

Explanation:
1. Zinc removes the two vicinal bromines to give the alkene 3-methylpent-2-ene
2. HI with red phosphorus reduces the alkene to the alkane
3. The product is 3-methylpentane

Teacher's Note:
1. Link each named reaction to its reagent and the change in carbon number.
2. Elimination gives alkenes, reduction gives alkanes, and coupling doubles the alkyl chain.

Answer: (d) Ethane

Explanation:
1. Decarboxylation of \( \ce{CH3CH2COONa} \) removes the carboxyl carbon
2. Ethane is formed

Teacher's Note:
1. Identify the functional group that is removed or formed and count the carbon atoms of the product.
2. Wurtz and Kolbe couple two alkyl groups, soda lime removes one carbon, and Grignard reagents pick up an acidic proton.